<p>You are given a string <code>s</code> consisting of digits and an integer <code>k</code>.</p>

<p>A <strong>round</strong> can be completed if the length of <code>s</code> is greater than <code>k</code>. In one round, do the following:</p>

<ol>
	<li><strong>Divide</strong> <code>s</code> into <strong>consecutive groups</strong> of size <code>k</code> such that the first <code>k</code> characters are in the first group, the next <code>k</code> characters are in the second group, and so on. <strong>Note</strong> that the size of the last group can be smaller than <code>k</code>.</li>
	<li><strong>Replace</strong> each group of <code>s</code> with a string representing the sum of all its digits. For example, <code>&quot;346&quot;</code> is replaced with <code>&quot;13&quot;</code> because <code>3 + 4 + 6 = 13</code>.</li>
	<li><strong>Merge</strong> consecutive groups together to form a new string. If the length of the string is greater than <code>k</code>, repeat from step <code>1</code>.</li>
</ol>

<p>Return <code>s</code> <em>after all rounds have been completed</em>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;11111222223&quot;, k = 3
<strong>Output:</strong> &quot;135&quot;
<strong>Explanation:</strong> 
- For the first round, we divide s into groups of size 3: &quot;111&quot;, &quot;112&quot;, &quot;222&quot;, and &quot;23&quot;.
  ​​​​​Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. 
&nbsp; So, s becomes &quot;3&quot; + &quot;4&quot; + &quot;6&quot; + &quot;5&quot; = &quot;3465&quot; after the first round.
- For the second round, we divide s into &quot;346&quot; and &quot;5&quot;.
&nbsp; Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. 
&nbsp; So, s becomes &quot;13&quot; + &quot;5&quot; = &quot;135&quot; after second round. 
Now, s.length &lt;= k, so we return &quot;135&quot; as the answer.
</pre>

<p><strong>Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;00000000&quot;, k = 3
<strong>Output:</strong> &quot;000&quot;
<strong>Explanation:</strong> 
We divide s into &quot;000&quot;, &quot;000&quot;, and &quot;00&quot;.
Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. 
s becomes &quot;0&quot; + &quot;0&quot; + &quot;0&quot; = &quot;000&quot;, whose length is equal to k, so we return &quot;000&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>2 &lt;= k &lt;= 100</code></li>
	<li><code>s</code> consists of digits only.</li>
</ul>
